answer:[Solution] The formula for the sum of the first n terms of an arithmetic sequence S_{n} is2 S_{n}=n[2 a+(n-1) d], where the first term is a, and the common difference is d.From the problem, we havebegin{array}{c}200=10(2 a+9 d) text { and } quad 20=100(2 a+99 d), 2 S_{110}=110(2 a+109 d) .end{array}From the first two equations, we get quad 2 a+109 d=-2.Therefore, 2 S_{110}=110 cdot(-2), so S_{110}=-110. Hence, the answer is (D).

answer:Impose a coordinate system on the diagram where point D is the origin. Therefore A=(0,42), B=(84,42), C=(84,0), and D=(0,0). Because M is a midpoint and N is a trisection point, M=(0,21) and N=(28,42). The equation for line DN is y=frac{3}{2}x and the equation for line CM is frac{1}{84}x+frac{1}{21}y=1, so their intersection, point O, is (12,18). Using the shoelace formula on quadrilateral BCON, or drawing diagonal overline{BO} and using frac12bh, we find that its area is 2184. Therefore the area of triangle BCP is frac{2184}{2} = 1092. Using A = frac 12 bh, we get 2184 = 42h. Simplifying, we get h = 52. This means that the x-coordinate of P = 84 - 52 = 32. Since P lies on frac{1}{84}x+frac{1}{21}y=1, you can solve and get that the y-coordinate of P is 13. Therefore the area of CDP is frac{1}{2} cdot 13 cdot 84=boxed{546}.

answer:First, we need to split the integral into two parts at x=2, where the absolute value function changes its behavior. Therefore, we have:int_{0}^{4}|x-2|dx = int_{0}^{2}(2-x)dx + int_{2}^{4}(x-2)dx.Now, let's evaluate each integral separately.For the first integral, we have:int_{0}^{2}(2-x)dx = left[2x - frac{1}{2}x^2right]_0^2 = (4 - 2) - (0 - 0) = 2.For the second integral, we have:int_{2}^{4}(x-2)dx = left[frac{1}{2}x^2 - 2xright]_2^4 = (8 - 8) - (2 - 4) = 2.Finally, adding the two results together, we obtain:int_{0}^{4}|x-2|dx = 2 + 2 = boxed{4}.

answer:To simplify and then evaluate the given expression [left(3x+2yright)left(3x-2yright)-left(3x-2yright)^{2}]div left(4yright) with x=frac{1}{2} and y=-frac{1}{4}, we follow these steps:1. Simplify the expression inside the brackets: begin{align*} [left(3x+2yright)left(3x-2yright)-left(3x-2yright)^{2}] &= [9x^{2}-4y^{2}-(9x^{2}-12xy+4y^{2})] &= 9x^{2}-4y^{2}-9x^{2}+12xy-4y^{2} &= 12xy-8y^{2} end{align*}2. Divide the simplified expression by (4y): begin{align*} (12xy-8y^{2})div left(4yright) &= frac{12xy}{4y} - frac{8y^{2}}{4y} &= 3x - 2y end{align*}3. Substitute (x=frac{1}{2}) and (y=-frac{1}{4}) into the simplified expression: begin{align*} 3x - 2y &= 3left(frac{1}{2}right) - 2left(-frac{1}{4}right) &= frac{3}{2} + frac{1}{2} &= 2 end{align*}Therefore, after simplifying and substituting the given values of (x) and (y), the final answer is boxed{2}.

answer:Given the equations xy=-2 and x+y=4, we aim to find the value of the expression x^{2}y+xy^{2}.Starting with the given expression:[x^{2}y+xy^{2}]We can factor out the common term xy:[= xy(x+y)]Substituting the given values for xy and x+y:[= (-2) times 4]Performing the multiplication:[= -8]Therefore, the value of the algebraic expression x^{2}y+xy^{2} is boxed{-8}.

answer:(1) If the complex number z is a real number, then m^2-1=0, so m=pm1; (2) If the complex number z is an imaginary number, then m^2-1neq 0, so mneqpm1; (3) If the complex number z is a pure imaginary number, then begin{cases} m^2+m-2=0 m^2-1neq 0 end{cases}, so m=-2.Therefore, the answers are:(1) m=pm1 for z to be a real number, so boxed{m=pm1}; (2) mneqpm1 for z to be an imaginary number, so boxed{mneqpm1}; (3) m=-2 for z to be a pure imaginary number, so boxed{m=-2}.